Using implicit differentiation find $\\displaystyle \\frac{dy}{dx}$ in terms of $x$ and $y$.
\nInput your answer here:
\n$\\displaystyle \\frac{dy}{dx}= $ [[0]]
\n \n \n ", "variableReplacementStrategy": "originalfirst", "steps": [{"prompt": "To differentiate implictly, we must differentiate both sides of the equation with respect to $x$
\nNotice (from the chain rule) that $\\frac{d}{dx}f(y) = \\frac{df}{dy}\\frac{dy}{dx}$
\nHence, for example $\\frac{d}{dx}y^2 = \\frac{d}{dy}(y^2)\\frac{dy}{dx}=2y\\frac{dy}{dx}$
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\n \t\t \t\t \t\tGiven $x^2+y^2+ax+by=c$ find $\\displaystyle \\frac{dy}{dx}$ in terms of $x$ and $y$.
\n \t\t \t\t \t\t\n \t\t \t\t \n \t\t \n \t\t"}, "name": "Simon's copy of SFY0004 Implicit 1", "statement": "
Given the following relation between $x$ and $y$
\\[\\simplify[all,!collectNumbers]{x^2+y^2+{a}x+{b}y}=\\var{c}\\]
answer the following question.
To differentiate implictly, we must differentiate both sides of the equation with respect to $x$
\nNotice (from the chain rule) that $\\frac{d}{dx}f(y) = \\frac{df}{dy}\\frac{dy}{dx}$
\nHence, for example $\\frac{d}{dx}y^2 = \\frac{d}{dy}(y^2)\\frac{dy}{dx}=2y\\frac{dy}{dx}$
\n\n\n(a)
\nOn differentiating both sides of the equation implicitly we get
\\[2x + \\simplify[all,!collectNumbers]{2y*Diff(y,x,1) + {a} + {b} *Diff(y,x,1)} = 0\\]
Collecting terms in $\\displaystyle\\frac{dy}{dx}$ and rearranging the equation we get
\\[(\\var{b} + 2y) \\frac{dy}{dx} = \\simplify[all,!collectNumbers]{{ -a} -2x}\\] and hence on further rearranging:
\\[\\frac{dy}{dx} = \\simplify[all,!collectNumbers]{({ - a} - 2 * x) / ({b} + (2 * y))}\\]